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umka2103 [35]
3 years ago
12

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

it reaches the bottom of the ramp
Physics
1 answer:
Arturiano [62]3 years ago
4 0

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s

So, its velocity at the bottom of the ramp is 2.21 m/s.

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A father pushes his child in a cart. The cart starts to move.
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The potential energy of the car when it let go is 20,000 J.

The speed of the car at the bottom of the ramp is 20 m/s.

The given parameters;

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<em />

The potential energy of the car is calculated as follows;

P.E = mgh

P.E = 100 x 10 x 20

P.E = 20,000 J

The speed of the car at the bottom of the ramp is calculated as follows;

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Samburu the elephant has a mass of 1650kg. If each of his feet cover an area of .25 m^2, how much pressure does samburu exert on
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Samburu's weight is (mass x gravity) = (1650kg x 9.8 m/s²) = 16,170 Newtons.

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Pressure = (force) / (area) = (16,170 Newtons) / (1 m²) = 16,170 Pascal .
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